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  1. Practice
  2. Tutorials
  3. 30 Days of Code
  4. Day 19: Interfaces
  5. Discussions

Day 19: Interfaces

  • dropyghost + 3 comments

    at least you could stop checking at n/2 and add n at the end

    • Inkling + 3 comments

      Yeah it's a very easy optimization, here's how mine looked:

      class Calculator : AdvancedArithmetic {
  
  public int divisorSum(int n) {
    int sum = n;
    for (int i = 1, half = n/2; i <= half; i++) {
      if (n % i == 0) sum += i;
    }
    return sum;
  }
  
}
      • raviloga316 + 0 comments
        [deleted]
      • nuhman + 0 comments

        That's clever

      • dashingakd + 0 comments

        u diid't declare half?

    • wy218032 + 0 comments
      [deleted]
    • wy218032 + 1 comment
      //we could stop at sqrt(n)
//This is my Java solution. I think this may be better.

class Calculator implements AdvancedArithmetic{
    public int divisorSum(int n){
       int sum = 0;
       for(int i = 1; i <= (int)Math.sqrt(n); i++) {
           if(n%i==0){
               sum = sum + i + n/i;
           }
       }
       if((int)Math.sqrt(n)*(int)Math.sqrt(n)==n){
           sum -= Math.sqrt(n);
       }
        return sum;
    }
}
      • nuhman + 2 comments

        I din't understand why you are doing the

        if((int)Math.sqrt(n)*(int)Math.sqrt(n)==n){
           sum -= Math.sqrt(n);
       }

        part. Can you explain?

        • impex_ua + 0 comments

          If n is square number, you have to minus sqrt n. Just math.

        • toraju + 1 comment

          You can optimize that one step to the following... to avoid correction for doubble addition, when input number is an exact square.

          public int divisorSum(int n){
       int sum = 0;int i;
       for( i = 1; i <(int)Math.sqrt(n); i++) {
           if(n%i==0){
               sum = sum + i + n/i;
           }
       }
       if(i*i == n){
           sum = sum+i;
       }
        return sum;
    }
          • Ankurjain04 + 0 comments

            sum = sum+i; should be sum = sum-i;